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Math Help - Proving identities, again.

  1. #1
    Member Furyan's Avatar
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    Proving identities, again.

    Iím sorry to ask for help again , but I canít prove the following identity:

    \dfrac{\tan^3\theta}{1 + \tan^2\theta} + \dfrac{\cot^3\theta}{1 + \cot^2\theta} \equiv \dfrac{1 -2\sin^2\theta\cos^2\theta}{\sin\theta\cos\theta}

    I have tried but Iím afraid I simply canít do it.

    I keep getting:

    LHS \equiv \dfrac{\sin^4\theta + cos^4\theta}{\sin\theta\cos\theta}

    And I canít see how to equate the numerators. I think I've probably made a mistake.

    A hint would be very much appreciated.

    Thank you.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Proving identities, again.

    Note that:
    \sin^4(\theta)+\cos^4(\theta)=\sin^4(\theta)+[1-\sin^2(\theta)]^2=\ldots
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  3. #3
    Member Furyan's Avatar
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    Re: Proving identities, again.

    Thank you Siron

    Quote Originally Posted by Siron View Post
    Note that:
    \sin^4(\theta)+\cos^4(\theta)=\sin^4(\theta)+[1-\sin^2(\theta)]^2=\ldots
    That's brilliant. It so simple but I just couldn't see it, which is very frustrating. I'll work through that and see if I can do it.

    Thank you.
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  4. #4
    Member Furyan's Avatar
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    Re: Proving identities, again.

    Quote Originally Posted by Siron View Post
    Note that:
    \sin^4(\theta)+\cos^4(\theta)=\sin^4(\theta)+[1-\sin^2(\theta)]^2=\ldots
    That wasn't as easy as I thought it was going to be. My instinct was to write \sin^4\theta = [1 - \cos^2\theta]^2 as well, but that was a mess. I got there, following your work, by expanding and factoring the expression you gave me.

    Thank you very much. There's so much to learn and you are all really helping a lot.
    Last edited by Furyan; December 24th 2011 at 10:09 AM. Reason: Correction
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Proving identities, again.

    We have:
    \sin^4(\theta)+[1-\sin^2(\theta)]^2=\sin^4(\theta)+1-2\sin^2(\theta)+\sin^4(\theta)=1-2\sin^2(\theta)+2\sin^4(\theta)=1-2\sin^2(\theta)[1-\sin^2(\theta)]=1-2\sin^2(\theta)\cos^2(\theta)
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  6. #6
    Member Furyan's Avatar
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    Re: Proving identities, again.

    Quote Originally Posted by Siron View Post
    We have:
    \sin^4(\theta)+[1-\sin^2(\theta)]^2=\sin^4(\theta)+1-2\sin^2(\theta)+\sin^4(\theta)=1-2\sin^2(\theta)+2\sin^4(\theta)=1-2\sin^2(\theta)[1-\sin^2(\theta)]=1-2\sin^2(\theta)\cos^2(\theta)
    Thank you. Yes, that's how I got there in the end.



    Only one to go, then integration with trig. functions, yipee!
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  7. #7
    MHF Contributor Siron's Avatar
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    Re: Proving identities, again.

    You're welcome!
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